A uniform circular metal disk has radius 35.0 cm and mass 350.0 g and its center

Question

A uniform circular metal disk has radius 35.0 cm and mass 350.0 g and its center is at the origin. Then a circular hole of radius 11.667 cm is cut out of it. The center of the hole is a distance 17.5005 cm from the center of the disk. The disk is rotating at a rate of 2.2 rad/sec.

1) What is the moment of inertia of the modified disk about the origin?

2) What is the total linear velocity of a piece metal at the disk's outer edge?

3) What is the total linear velocity of a piece metal 10.167 cm from the center?

Explanation / Answer

A circular steel plate disk of radius 1.9 m, thickness 0.04 m, and density 8477.0 kg/m2 has a circular hole of radius 0.5 m located halfway between the center and circumference of the large steel disk. Find the distance of the center of mass of the steel disk from its center. How far is the center of mass from the center of the hole? I am not sure where to even start. I attemped to compute the problem by using the center of mass equation we recieved in class, but I got a negative number which can't be correct. A negative number is fine. You can expect, intuitively, that if the hole is a little bit to the right (in the positive-x direction), then the CM is going to be a little bit to the left (in the negative-x direction). In general, the CM is a particular point, which may very well have negative coordinates. Since you're asked only to find the DISTANCE from the center, take the absolute value of the CM's coordinate. To calculate the CM of a simple geomentric figure with a "hole" in it, it's very useful to remember this about centers of mass: Given two objects "a" and "b" whose CM's and masses are known, you can calculate the CM of the assembly (ab) as follows: CM_ab = [(m_a)(CM_a) + (m_b)(CM_b)] / (m_a + m_b) In this case, instead of ADDING two pieces together, we're going to SUBTRACT one piece from another (i.e. subtract a small plug of steel from the large disk). But we can still use the same equation. So: start by aligning the large disk so its center is at the origin of the x-y plane, and the center of the "hole" is on the positive x-axis. Let the large disk (with its hole) be object "a". For object "b", we'll use an imaginary small steel disk that can exactly fill the hole. So, imagine "assembly ab" equals the large disk "a" with the small disk "b" plugging up the hole. By the equation above: CM_ab = [(m_a)(CM_a) + (m_b)(CM_b)] / (m_a + m_b) But clearly, the CM of the "assembly" is right in the center of the large disk, since the "assembly" is just a uniform disk. So that means CM_ab = 0. 0 = [(m_a)(CM_a) + (m_b)(CM_b)] / (m_a + m_b) which means: (m_a)(CM_a) = -(m_b)(CM_b) or: CM_a = -(CM_b)(m_b)/(m_a) Since the center of the "plug" is given as "halfway between the center and circumference of the large steel disk," we know that CM_b equal 0.25 meters: CM_a = -(0.25 meters)(m_b)/(m_a) Also, since the disk and the "plug" are of the same density, the ratio of their masses is as the ratio of their volumes. And when two disks have the same thickness, the ratio of their volumes is as the ratio of the square of their radii (you can easily convince yourself of this by calculating it). So we have: CM_a = -(0.25 meters)(r_b)²/(r_a)² = -(0.25 meters)(0.5 meters)²/(1.9 meters)²

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